Question

An urn contains 12 balls, of which 4 are white.

Answer 1

Answer:

a) $\frac{1}{3}$

b) $\frac{4}{12}\,,\,\frac{4}{11}\,,\,\frac{4}{10}\,,...$

Step-by-step explanation:

Number of balls in the urn = 12

Number of white balls in the urn = 4

So, number of balls which are not white = 12 - 4 = 8

We know that probability = No. of outcomes/Total number of outcomes

Let $A_i\,,\,B_i\,,\,C_i$ denotes the event of getting white ball by A, B and C

a) each ball is replaced after being drawn.

Probability = Number of white balls/Total number of balls

Solution: $P(A_i)=P(B_i)=P(C_i)=\frac{4}{12}=\frac{1}{3}$

b)the balls that are withdrawn are not replaced.

Solution:

If A wins: $P(A_1)=\frac{4}{12}=\frac{1}{3}$

If A lost and B wins: $P\left ( B_1|A_1^{c} \right )=\frac{4}{11}$

If A and B lost and C wins: $P\left ( C_1|A_1^{c}\,B_1^{c} \right )=\frac{4}{10}$

and so on....